Problems & Puzzles: Puzzles

Puzzle 674 Concatenating the squares of consecutive primes Suppose that we concatenate the square of k consecutive primes p1, p2,... pk: C=p1^2&p2^2...pk^2 Q1. For what combinations of P1 & k  C is always composite? Q2. Send your largest prime C and appropriate conditions for p1=3 & k.

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Contributions came from Seiji Tomita, Jan van Delden & Emmanuel Vantieghem.

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Seiji wrote:

Q1.
Case1. p1<>3
Generally pk^2=1 mod 3 for pk <> 0 mod 3.
If k=0 mod 3, then c is always divisible by 3.

Case2. p1=3
If k=1 mod 3, then c is always divisible by 3.

Q2.
c=3^2&5^2&7^2&...&883^2 is a prime number.
 

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Jan wrote:

Q1  

 

I see no general condition on (p1,K) such that C(p1,k) is composite for k<=K (for instance).

 

Since for all primenumbers p (except 3) we have p^2=1 mod 3 and since 10=1 mod 3, we have

3|C(3,k) if k=1 mod 3 and 3|C(p,k) if k=0 mod 3, p<>3.

 

Maybe some similar identities exist, I didn’t pursue this any further.

 

Q2   

 

The smallest K for which C(3,k) is composite for k<=K and prime for k=K+1 equals K=151.
With prime C(3,152)=

9254912116928936152984196113691681184922092809348137214489504153296241\

6889792194091020110609114491188112769161291716118769193212220122801246\

4926569278892992932041327613648137249388093960144521497295152952441542\

8957121580816300166049691697236173441767297896180089858499424996721979\

6910048910956111356912040912180112460912888113468913912914364114668915\

1321157609160801167281175561177241185761187489192721196249201601208849\

2125212143692180892294412371692410812490012530092590812714412735292926\

8129920931024931696932376132604133292934456935164935880136120136844937\

5769380689383161398161410881413449418609426409434281436921452929458329\

4664894774814914015026815169615285295372895461215520495640015730495791\

2159136159752961936963520965448165772167404167732968392968724170392172\

7609734449737881744769769129776161779689

having 810 digits.

 

The next “prime” is: C(3,2340) and has 19118 digits. It is prime with probability at least 1-(1/2)^100. It is a bit too large to write down (I think) and should be verified, considering it’s size, the last 9 digits are 431517529, the last prime used is therefore 20773, which should be p[2341].

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Emmanuel wrote:

This is my best prime :

   C = (7^2)&(11^2)&...&(2417^2)  (2234 digits), proved prime by PRIMO in about 11 hours.

These are bigger ones :

  (37^2)&(41^2)&...&(5807^2)     (5348 digits)

  (11^2)&(13^2)&...&(7757^2)     (7145 digits)

but their primality proof requires too much computing time for me...

 

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